Question: $h(t) = -3t^{2}-4t+2(g(t))$ $g(t) = -3$ $f(n) = 6n+3(g(n))$ $ g(f(-6)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(-6)$ . Then we'll know what to plug into the outer function. $f(-6) = (6)(-6)+3(g(-6))$ To solve for the value of $f$ , we need to solve for the value of $g(-6)$ $g(-6) = -3$ $g(-6) = -3$ That means $f(-6) = (6)(-6)+(3)(-3)$ $f(-6) = -45$ Now we know that $f(-6) = -45$ . Let's solve for $g(f(-6))$ , which is $g(-45)$ $g(-45) = -3$